Difference between revisions of "1997 AHSME Problems/Problem 9"
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<math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3-\frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \frac{11}{5}\qquad\textbf{(D)}\ \sqrt{5}\qquad\textbf{(E)}\ \frac{9}{4} </math> | <math> \textbf{(A)}\ 2\qquad\textbf{(B)}\ 3-\frac{\sqrt{3}}{2}\qquad\textbf{(C)}\ \frac{11}{5}\qquad\textbf{(D)}\ \sqrt{5}\qquad\textbf{(E)}\ \frac{9}{4} </math> | ||
+ | |||
+ | ==Solution 1== | ||
+ | |||
+ | Since <math>\angle EBA = \angle FCB</math> and <math>\angle FBC = \angle AEB</math>, we have <math>\triangle ABE \sim \triangle FCB</math>. | ||
+ | |||
+ | <math>\frac{AB}{FC} = \frac{BE}{CB} = \frac{EA}{BF}</math> | ||
+ | |||
+ | <math>\frac{2}{FC} = \frac{\sqrt{5}}{2} = \frac{1}{BF}</math> | ||
+ | |||
+ | From those two equations, we find that <math>CF = \frac{4}{\sqrt{5}}</math> and <math>BF = \frac{2}{\sqrt{5}}</math> | ||
+ | |||
+ | Now that we have <math>BF</math> and <math>CF</math>, we can find the area of the bottom triangle <math>\triangle CFB</math>: <math>\frac{1}{2}\cdot \frac{4}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = \frac{4}{5}</math> | ||
+ | |||
+ | The area of left triangle <math>\triangle BEA</math> is <math>\frac{1}{2}\cdot 2 \cdot 1 = 1</math> | ||
+ | |||
+ | The area of the square is <math>4</math>. | ||
+ | |||
+ | Thus, the area of the remaining quadrilateral is <math>4 - 1 - \frac{4}{5} = \frac{11}{5}</math>, and the answer is <math>\boxed{C}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Place the square on a coordinate grid so that <math>B(0,0)</math> and <math>D(2,2)</math>. Line <math>BE</math> is <math>y = 2x</math>. Line <math>CF</math> goes through <math>(2,0)</math> and has slope <math>-\frac{1}{2}</math>, so it must be <math>y = 1 -\frac{x}{2}</math> | ||
+ | |||
+ | The intersection of the two lines is <math>F</math>, and <math>F</math> thus has coordinates <math>(\frac{2}{5}, \frac{4}{5})</math>. The altitude from <math>F</math> to <math>BC</math> thus has length <math>\frac{4}{5}</math>, so the area of the triangle <math>BCF</math> is <math>\frac{1}{2}\cdot 2\cdot \frac{4}{5} = \frac{4}{5}</math>. | ||
+ | |||
+ | The other triangle has area <math>1</math>, and the whole square has area <math>4</math>. As above, we find the area of the quadrilateral by subtracting the two triangles, and we get <math>\frac{11}{5}</math>, which is <math>\boxed{C}</math>. | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1997|num-b=8|num-a=10}} | ||
+ | {{MAA Notice}} |
Latest revision as of 14:12, 5 July 2013
Contents
Problem 9
In the figure, is a
square,
is the midpoint of
, and
is on
. If
is perpendicular to
, then the area of quadrilateral
is
Solution 1
Since and
, we have
.
From those two equations, we find that and
Now that we have and
, we can find the area of the bottom triangle
:
The area of left triangle is
The area of the square is .
Thus, the area of the remaining quadrilateral is , and the answer is
Solution 2
Place the square on a coordinate grid so that and
. Line
is
. Line
goes through
and has slope
, so it must be
The intersection of the two lines is , and
thus has coordinates
. The altitude from
to
thus has length
, so the area of the triangle
is
.
The other triangle has area , and the whole square has area
. As above, we find the area of the quadrilateral by subtracting the two triangles, and we get
, which is
.
See also
1997 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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